# Steam enters a nozzle with a pressure of 10 MPa and a temperature of 400 degrees Celsius. The...

## Question:

Steam enters a nozzle with a pressure of 10 MPa and a temperature of 400 degrees Celsius. The steam leaves the nozzle at a pressure of 0.20 MPa.

Determine the exit temperature of the steam if the process in the nozzle is isentropic.

## Isentropic Process:

In an isentropic process, the entropy remains constant. Therefore {eq}\displaystyle \Delta S {/eq} = 0. All reversible adiabatic processes are isentropic. But in the real world, most thermodynamics processes, undergo some losses and therefore are irreversible.

## Answer and Explanation:

Inlet conditions of steam are 10 MPa and 400°C

From the steam table,

Enthalpy at inlet, {eq}\displaystyle h_1 = 3100 \ kJ/kg {/eq}

Entropy at inlet, {eq}\displaystyle s_1 = 6.218 \ kJ/kgK {/eq}

The process is isentropic. Therefore, entropy remains constant through the process.

{eq}\displaystyle s_2 = s_1 = 6.218 \ kJ/kgK {/eq}

Pressure at exit = 0.2 MPa

From the steam tables, at a saturation pressure of 0.2 MPa, the saturated vapor entropy, {eq}\displaystyle s_g {/eq} is 7.127 kJ/kgK and the saturated liquid entropy, {eq}\displaystyle s_f {/eq} is 1.530 kJ/kgK.

Therefore, the steam at exit of the nozzle is in a wet vapor state, since the entropy is lesser than the entropy of saturated vapor at that pressure and greater than the entropy of saturated liquid at that pressure.

At wet vapor state, the temperature of the steam is saturation temperature corresponding to 0.2 MPa which is 120.2°C.

#### Learn more about this topic:

Thermodynamics Practice Problems & Solutions

from Physics 112: Physics II

Chapter 3 / Lesson 6
7.6K